Lời giải:

Vi $a,b,c>0\Rightarrow a+b+c\neq 0$

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\)

\(\Rightarrow \frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=2\)

\(\Rightarrow \left(\frac{a+b}{c}\right)^3+\left(\frac{b+c}{a}\right)^3+\left(\frac{c+a}{b}\right)^3=2^3+2^3+2^3=24\) (đpcm)

Lời giải:

Ta có: \(\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\)

Mà: a = b + c => c = a - b => \(\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\)

=\(\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left[a^2-a\left(a-b\right)+\left(a-b\right)^2\right]}\)

\(=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left[a^2-a^2+ab+\left(a^2-2ab+b^2\right)\right]}\)

= \(\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-a^2+ab+a^2-2ab+b^2\right)}\)

\(=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ab+b^2\right)}=\frac{a+b}{a+c}\)

Vây: \(\frac{a^3+b^3}{a^3+c^3}=\frac{a+b}{a+c}\)

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a/

\(a^2+b^2+c^2+29ab+bc+ca=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Rightarrow a=b=c\)

b/ \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b\right)\)

\(=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)

c/ Không, vì \(a=b=c\ne\) thì \(a^3+b^3+c^3=3a^3=3abc\) vẫn đúng

1. \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)

\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

2. \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

3.Còn có a + b + c = 0 nữa mà bn.

\(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)

+ \(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

Bài 6 . Áp dụng BĐT Cauchy , ta có :

a² + b² ≥ 2ab ( a > 0 ; b > 0)

⇔ ( a + b)² ≥ 4ab

⇔ \(\dfrac{\left(a+b\right)^2}{4}\)≥ ab

⇔ \(\dfrac{a+b}{4}\) ≥ \(\dfrac{ab}{a+b}\) ( 1 )

CMTT , ta cũng được : \(\dfrac{b+c}{4}\) ≥ \(\dfrac{bc}{b+c}\) ( 2) ; \(\dfrac{a+c}{4}\) ≥ \(\dfrac{ac}{a+c}\)( 3)

Cộng từng vế của ( 1 ; 2 ; 3 ) , Ta có :

\(\dfrac{a+b}{4}\) + \(\dfrac{b+c}{4}\) + \(\dfrac{a+c}{4}\) ≥ \(\dfrac{ab}{a+b}\) + \(\dfrac{bc}{b+c}\) + \(\dfrac{ac}{a+c}\)

⇔ \(\dfrac{a+b+c}{2}\) ≥ \(\dfrac{ab}{a+b}\) + \(\dfrac{bc}{b+c}\) + \(\dfrac{ac}{a+c}\)

Bài 4.

Áp dụng BĐT Cauchy cho các số dương a , b, c , ta có :

\(1+\dfrac{a}{b}\) ≥ \(2\sqrt{\dfrac{a}{b}}\) ( a > 0 ; b > 0) ( 1)

\(1+\dfrac{b}{c}\) ≥ \(2\sqrt{\dfrac{b}{c}}\) ( b > 0 ; c > 0) ( 2)

\(1+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{c}{a}}\) ( a > 0 ; c > 0) ( 3)

Nhân từng vế của ( 1 ; 2 ; 3) , ta được :

\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\) ≥ \(8\sqrt{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=8\)

@Giáo Viên Hoc24.vn

@Akai Haruma

Biến đổi :

\(VT=\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}=\frac{a}{b\left(a+b^2\right)}+\frac{b}{c\left(b+c^2\right)}+\frac{c}{a\left(c+a^2\right)}\)

\(=\frac{1}{b}\cdot\frac{a}{a+b^2}+\frac{1}{c}\cdot\frac{b}{b+c^2}+\frac{1}{a}\cdot\frac{1}{c+a^2}\)

\(=\frac{1}{b}\cdot\left(1-\frac{b^2}{a+b^2}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{b+c^2}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{c+a^2}\right)\)

Áp dụng BĐT Cô-si :

\(VT\ge\frac{1}{b}\cdot\left(1-\frac{b^2}{2b\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{2c\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{2a\sqrt{c}}\right)\)

\(=\frac{1}{b}-\frac{1}{2\sqrt{a}}+\frac{1}{c}-\frac{1}{2\sqrt{b}}+\frac{1}{a}-\frac{1}{2\sqrt{c}}\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)

Áp dụng BĐT quen thuộc : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) và BĐT Cô-si ta có:

\(VT\ge\frac{9}{a+b+c}-\frac{1}{2}\cdot\left(\frac{\frac{1}{a}+1}{2}+\frac{\frac{1}{b}+1}{2}+\frac{\frac{1}{c}+1}{2}\right)\)

\(=\frac{9}{3}-\frac{1}{2}\cdot\left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}{2}\right)\ge3-\frac{1}{2}\cdot\left(\frac{\frac{9}{a+b+c}+3}{2}\right)\)

\(=3-\frac{1}{2}\cdot\left(\frac{\frac{9}{3}+3}{2}\right)=\frac{3}{2}\)

Ta có đpcm.

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Áp dụng hđt mở rộng đi bạn :)